Can I identify $X/A$ with the one-point compactification of $X-A$, where $X\supset A$ is a topological space? Intuitively, $A$ collapses to a single point which may represent the infinity point of the one point compactification of $X-A$. Definitely, we should assume $X$ is locally compact Hausdorff.
For example, if $X=S^n$ and $A={pt}$, a single point, then $X/A \cong X = S^n$ while $X-A \cong \mathbb R^n$, whose one point compactification is exactly $S^n$.
However, I am not sure if this is right for general topological spaces. And, I am trying to find out the exact conditions to guarantee this identification. Please help. Thanks.
 A: (Partial answer)
Edit: Corrected earlier omission: it is also necessary that $A$ be closed.
As G. Sassatelli points out, the definition of an open set in $X / A$ and in the compactification of $(X \setminus A)$ are different. A partial answer to your question is to determine under what conditions the canonical bijection between $X / A$ and the compactifiation of $(X \setminus A)$ is a homeomorphism.
First we consider the topology restricted to $(X \setminus A)$. $X \setminus A$ is open in the one-point compactification of the space $X \setminus A$, so it must also be open in $X / A$. It follows then immediately that $\boldsymbol{A}$ must be closed.
This guarantees that the topology on $X \setminus A$ has the same open sets as in $X$, so that open sets in $X \setminus A$ are the same as in the original topology of $X$ in both spaces.
From here, the additional condition we want is that the open sets coincide for sets containing the extra point: for any set $V \subseteq (X \setminus A)$, $V \cup \{[A]\}$ is open in $X/A$ if and only if $V \cup \{\infty\}$ is open in the compactification of $(X \setminus A)$. Unpacking these definitions, we want
$$
V \cup A \text{ is open} \iff X \setminus (V \cup A) \text{ is compact}
$$
The $\Longleftarrow$ is true if $X$ is Hausdorff: if $X \setminus (V \cup A)$ is compact, then it is closed, so its complement $V \cup A$ is open.
The $\Longrightarrow$ direction says: the complement of an open set containing $A$ is compact. Or in other words, a closed set disjoint from $A$ must be compact.
In summary:


*

*If $X$ is Hausdorff, then a necessary and sufficient condition is that (i) $A$ is closed, and (ii) every closed set disjoint from $A$ is compact. (Perhaps others can reduce this condition further to something else well-known.)

*In particular, if $X$ is compact Hausdorff and $A$ is closed, then the two are always equivalent. This covers the case you discussed, of $S^n$.
